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Basics of Optical Imaging in Microlithography: A "Hands-on" Approach

Tom D. Milster (University of Arizona) Robert Socha (ASML) Peter Brooker (SYNOPSYS)

Thanks to: • Del Hansen • Phat Lu •Warren Bletscher Milster, Socha, Brooker SPIE- SC707 1

What we want to do with this course

From This

Source Aperture Grating (Mask) Condenser Lens 1 Stop

To This

Image Plane (Aerial Image of Mask) Lens 2 CCD Camera (AIMS)

• • • •

f c f c f 1 f 1 f 2 f 2 Take a complicated optical system, like a lithographic projection camera used to make computer chips, and simplify it to a working model that demonstrates basic principles.

2f cam Use a simple optical system for the student to work with “hands on” and observe the results.

Demonstrate the relationship of the simple system to a real lithographic system through a commercial simulator.

Have fun and demonstrate our unparalled acting abilities 2f cam

Milster, Socha, Brooker SPIE- SC707 2

• • • • • •

OUTLINE

Intro

– – –

Basic Imaging – What we do in lithography The goal of making a small image What limits the size of the image?

Basic Illumination and Imaging

– –

Koehler Illumination Definition of coherence factor “sigma” Binary Mask

– – – –

Contrast versus pitch for sigma ~ 0 Contrast versus pitch for sigma > 0 2-Beam and 3-Beam Imaging Focus behavior Phase Mask

– –

Contrast versus pitch Focus behavior Off-Axis Illumination

– –

Contrast versus pitch Focus behavior Summary

Milster, Socha, Brooker SPIE- SC707 3

•

Introduction

What is photo lithography ?

Object: reticle or mask

• •

Optical image is recorded in the resist via changes in concentrations of species.

Concentration level controls development Optics

Photoresist Wafer + films Photoresist Development Aerial Image Latent Image

z y Resist Cross sections X

Negative Photoresist Milster, Socha, Brooker SPIE- SC707 Positive Photoresist

Etymology: Photolithography = Light Stone Writing

Introduction

• •

1st approximation is that Aerial image propagates into photoresist normal to the wafer plane, creating a latent image Reality is more complicated; you need to calculate E fields in photoresist at many propagation angles 0.25

m 5-BAR Structures Focus=0.0

m, NA=0.57 NA=0.6, 248nm

Z Image Cross Section Z Resist Cross Section (not top down!) Milster, Socha, Brooker SPIE- SC707 5

•

Introduction

The goal of making a small image

–

Transfer image into a photosensitive material, i.e., photoresist, for subsequent processing that results in a desired pattern to be used as a “stencil” photoresist

Milster, Socha, Brooker SPIE- SC707 6

•

Introduction

Imaging Resolution and Lord Rayleigh

–

Q: When can you resolve the image of 2 distance stars?

–

A: When the 1 st Intensity min of one lines up with peak of other

Small NA Large  Milster, Socha, Brooker SPIE- SC707

Resolution



0.61

 

From the math of the Airy function Large NA Web Top Optics, 1999 7

• • • •

Oh Master Litho… ..what limits the size of the photoresist pattern? Grasshopper, there are three paths to improve resolution:

•

Reduce Wavelength (Lambda)

• •

Increase numerical aperture (NA) Decrease k1 : “Process” knob

–

Includes off-axis illumination, complex masks, high contrast photoresist, acid diffusion, etc… ….now go away Grasshopper I am busy.

Milster, Socha, Brooker SPIE- SC707 8

• • •

What is it now Grasshopper… Master, what affects the contrast of the image?

The answer is found in the values of

•

CD and Pitch

•

Partial Coherence or illumination (s)

– –

s=0: Coherent Limit s=1: Incoherent Limit

Milster, Socha, Brooker SPIE- SC707 9

• • •

You again grasshopper… Master… …look at the following data

Milster, Socha, Brooker SPIE- SC707 10

Effect of Varying

s 

=193nm, NA=0.75

Dense Lines vs.

(circular) 150nm L/S 100nm L/S

•

Master, how come in one case increasing sigma is good (100nm L/S) and in the other case, increasing sigma is bad (200nm L/S)?

•

It depends on the amount of diffraction orders that are being collected by the lens…now go away!

Milster, Socha, Brooker SPIE- SC707 11

• • • •

Master, I am sure that your answers are correct but… …yes Grasshopper… But I find these facts confusing. What is sigma? How can in some cases a larger sigma be good and in other cases a larger sigma be bad? And what the heck is k1?

Master…I do not want only the answers…I want to understand…please help me understand master…

• • •

Grasshopper… you are finally asking the right question Go to the optical bench now… It holds the answer to your questions!!

Milster, Socha, Brooker SPIE- SC707 12

LED Source Source Aperture Grating (Mask) Condenser Lens 1 Stop Lens 2 Image Plane (Aerial Image of Mask) CCD Camera (AIMS) f c f c f 1 f 1 f 2 f 2 2f cam 2f cam

Milster, Socha, Brooker SPIE- SC707

First Light – Get An Image

Let’s do an experiment:

–

Set up the bench with:

• •

Pinhole Source Aperture Stop of 6.35 mm (1/4 in) diameter.

− Put in the L (25.2µm) pitch mask and observe the aerial image.

− The grating simulates a mask.

− The aerial image simulates what is used to expose the resist.

− In our system, the aerial image is reimaged onto a CCD camera, which is like an Aerial Image Measurement System (AIMS).

− Draw picture of the light pattern at the stop.

Milster, Socha, Brooker SPIE- SC707 That is what your image looks like Draw the light pattern at the stop here.

•

Basic Illumination and Imaging

Kohler Illumination Image of source Stop Source Aperture Mask Plane Condenser

Lens 1

Imaging Lens

Lens 2

Aerial Image

• •

Field Stop of Imaging Lens is Aperture stop condenser and vice versa Lithographic systems use Koehler illumination where the illumination

source aperture is imaged into the stop of the imaging lens.

Milster, Socha, Brooker SPIE- SC707 15

•

Basic Illumination and Imaging

Definition of Coherence Factor ‘Sigma’ Mask Plane Source Image Diameter Stop Diameter Source Pupil Edge (the “NA”) Condenser Imaging Lens

σ



Diameter of Source Image in Stop Diameter of Stop

Source Image View of Entrance Pupil with blank mask

Milster, Socha, Brooker SPIE- SC707 16

•

Simple Binary Mask

Model a Cr on quartz grating mask as an infinitely thin grating

Note: For 1:1 lines and spaces, P= 2 * LW LW = Line Width

SiO 2 E E-Field Diffraction Orders

-3 -2 -1 q 0 +1 +2 +3

Position Grating Equation:

sin( q )  

Pitch

Lens/Pupil -1 st 0 th +1 st

Milster, Socha, Brooker SPIE- SC707 17

Effect of Varying Pitch

•

Let’s do an experiment

–

Set up the bench with:

•

Pinhole Source

– •

Aperture Stop of 6.35 mm (1/4 in) diameter.

Use the S(8.4µm), M(12.6µm) and L(25.2µm) pitches of the mask and observe the effects in the image plane and at the stop.

– –

Draw the light pattern at the stop on the next page.

What is the relationship between the light pattern at the stop and the image?

–

What is the smallest pitch for which we can obtain an image?

–

This system is very similar to what would be observed if an on-axis laser beam was used to illuminate the mask. Therefore, we call this case

coherent imaging .

SPIE- SC707 –

Notice that the lines in the image are either completely resolved, or they are not. There is no ‘partially resolved’ case.

Effect of Varying Pitch

Draw the light pattern at the stop for the S(8.4 µm) grating. Draw the light pattern at the stop for the M(12.6 µm) grating. Draw the light pattern at the stop for the L(25.2 µm) grating. Milster, Socha, Brooker SPIE- SC707 19

•

Binary Mask and Diffraction Orders

Must have more than 1 order in pupil to have image modulation

+3 Pupil (stop) +1 o

Strong Image Modulation

-1 We see

diffraction orders

emanating from the mask that are necessary for imaging.

pupil +1 -3 NA=sin( q Max ) q Max q Max pupil o -1 +1 o

For 1:1 grating

Coherent limit

min  

NA LW

min   2 

NA P min

is the minimum pitch that is at the limit of resolution. Photoresist CD k 1 =1/2   1 

No Image just constant Irradiance

-1 Milster, Socha, Brooker SPIE- SC707 20

Milster, Socha, Brooker SPIE- SC707

Coffee Break

•

Time for the Late Shows new and exciting quiz game sensation.

•

Do you want to play:

– – –

Know your “Current events”?

Know your “Cuts of Beef’?

Know your “Optics Bench Basics”?

•

Know your Bench Basics! Excellent choice!!!

Milster, Socha, Brooker SPIE- SC707 22

Bench basics:

Source Aperture Grating (Mask) Condenser Lens 1 Stop Lens 2 Image Plane (Aerial Image of Mask) CCD Camera (AIMS)

• • • • •

f c f c f 1 f 1 f 2 f 2 2f cam Where is the Source Aperture relative to the condenser lens?

Is it: A: at minus infinity B: it refuses to reveal its location C: The source aperture is located at the front focus of the condenser lens 2f cam

•

Answer is C: The source aperture (effective source for the system) is located at the focus of the condenser lens. Collimated light from the LED illuminates the grating. Light from every part of the source aperture illuminates each point on the grating.

Milster, Socha, Brooker SPIE- SC707 23

Bench basics

Source Aperture Grating (Mask) Condenser Lens 1 Stop Lens 2 Image Plane (Aerial Image of Mask) CCD Camera (AIMS) f c f c f 1 f 1 f 2 f 2 2f cam 2f cam

• • • • •

Q: Where does the image of the Source Aperture appear?

Does it appear … A: only in the Borg space time continuum B: at the grating C: in the plane of the “Stop”.

•

Correct answer is C: The image of the Source Aperture appears in the plane of the stop.

Milster, Socha, Brooker SPIE- SC707 24

Bench basics

Source Aperture Grating (Mask) Condenser Lens 1 Stop Lens 2 Image Plane (Aerial Image of Mask) CCD Camera (AIMS)

• • • •

f c f c f 1 f 1 f 2 f 2 2f cam 2f cam Q: Collimated light from the Source Aperture illuminates the Grating. This is because….

A: The grating is not worthy of the sources “focused” attention B: The source is the grating…question is irrelevant C: Kohler Illumination of the grating averages out non uniformities in the source.

•

Answer is C

Milster, Socha, Brooker SPIE- SC707 25

•

Comedy writer’s strike…

•

No more multiple choice answers

•

Let’s continue to cement the concepts associated with the bench

Milster, Socha, Brooker SPIE- SC707 26

Bench Basics

Source Aperture Grating (Mask) Condenser Lens 1 Stop Lens 2 Image Plane (Aerial Image of Mask) CCD Camera (AIMS) f c f c f 1 f 1 f 2 f 2 2f cam

• •

Q: Where is the grating located with respect to Lens1?

A: The grating is located at the focus of lens 1.

2f cam

• •

Q: Where does the image of the grating appear?

A: The image of the grating appears at the “Image plane”

Milster, Socha, Brooker SPIE- SC707 27

Bench Basics:

Source Aperture Grating (Mask) Condenser Lens 1 Stop Lens 2 Image Plane (Aerial Image of Mask) CCD Camera (AIMS) f c f c f 1 f 1 f 2 f 2 2f cam 2f cam

• •

Q: If the Image occurs at the image plane, why is the microscope needed?

A: The image of the source at the image plane cannot be seen with the eye. The microscope is needed to magnify the image so it can be seen by your eye.

Milster, Socha, Brooker SPIE- SC707 28

Bench Basics: Grating off axis point

Grating (Mask) Stop Image Plane (Aerial Image of Mask) Lens 1 Lens 2 f 1 f 1 f 2 f 2

• •

Q: Look at the above picture. Estimate the vertical magnification?

~3.7

• •

How can the vertical magnification be decreased?

Decrease f2 but keep “Stop” at focus of Lens2.

Milster, Socha, Brooker SPIE- SC707 29

Connection back to real Scanner Optics

• •

Q: Where is the mask plane and image of the mask?

A: First plane on the left and last plane on right.

• •

Q: Can you find the stop in the lens column?

A: On the right side of center.

• •

Q: What is the magnification?

A: 4x demagnification.

Milster, Socha, Brooker SPIE- SC707 30

•

Effect of Varying Sigma

Let’s do an experiment

–

Set up the bench with:

• • •

Pinhole Source Aperture Stop of 6.35 mm diameter.

S(8.4µm) grating

–

Use the PH, 3.18mm (1/8 in) and 6.35mm (1/4in) diameter sources and observe the effect at the stop and at the image plane. Estimate

for each source.

– –

Draw the light pattern at the stop on the next page.

Is there a point where we can resolve the lines in the image?

– – –

By changing

, we are allowing more light through the stop that can interfere to form an image.

Not all of the light that is passed through the stop can interfere, thus giving us background light that reduces our contrast. The amount of background light is a function of the pitch, therefore the contrast is a function of the pitch.

This case is called

partially coherent imaging

, because of the dependence of the contrast on pitch.

Milster, Socha, Brooker SPIE- SC707 31

Effect of Varying Sigma

Draw the light pattern at the stop for the PH light source. Draw the light pattern at the stop for the 3.18mm diameter light source. Draw the light pattern at the stop for the 6.45mm diameter light source. Milster, Socha, Brooker SPIE- SC707 32

Contrast Curves versus Pitch & Sigma

• • •

Sigma=0.05 ---Coherent Sigma=0.5 -----Partially Coherent Sigma=1 ---Incoherent limit

Milster, Socha, Brooker SPIE- SC707 33

Modulation Transfer Function (MTF)

• •

Optics types love this plot!!!!

Can you find the Coherent frequency cut off?

Milster, Socha, Brooker SPIE- SC707 34

Binary Mask: Influence of Sigma

•

Pupil diagrams with Partial Coherence :

NA’ We must have at least 2 conjugate sources points in the pupil to form an image.

σ No imaging Imaging!!

•

Each source point is projected by the diffraction orders from the mask

– –

These will interfere with each other for a given source point need more than 1 for interference and hence image modulation

Milster, Socha, Brooker SPIE- SC707 35

Binary Mask: Sigma < 1

Resolution limit with 0<

<1 for a circular source

s •

No grating - just blank mask 0th order

•

Grating period at cut-off frequency

 

NA LW

min  2  

-1st 0th order

  limit  2

(1  s )

+1st

•

Grating period resolution limit at given

min  (1 s  ) 

For 1:1 grating,

min   s ) 

Milster, Socha, Brooker SPIE- SC707

-1st 0th order +1st

Binary Mask: Sigma = 1

Resolution limit with

s

1 for a circular source

•

No grating - just blank mask 0th order

•

Grating period at cut-off frequency

 

•

Grating period corresponds to incoherent cut-off

min  2  

Milster, Socha, Brooker SPIE- SC707 For 1:1 grating,

min   4 

-1st

  limit  4



min -1st 0th order

+1st

+1st 37

Milster, Socha, Brooker SPIE- SC707

Binary Mask,



=248nm, NA=0.63

Milster, Socha, Brooker SPIE- SC707 s

=0.05 &

= 0.7 for k1=0.5

Different cases for on axis, k1=0.5

• • • • • •

Assume circular, on axis illumination Assume dense L/S k1=0.5

– – 

Center of n=1 diffraction orders are at edge of lens



CD = LW = 0.5*Lambda/NA For 248nm illumination, NA=0.63

–

CD = 0.5*248nm/0.63 = 197nm



200nm L/S give k1=0.5

For 193nm illumination, NA=0.93

–

CD = 0.5*193nm/0.93 = 104nm



104nm L/S give k1=0.5

For 193nm illumination, NA = 1.2

–

CD = 0.5*193/1.2 = 80.4nm



80nm L/S gives k1 = 0.5

• •

Results above are only good for on axis illumination.

The usual off-axis case is different.

Milster, Socha, Brooker SPIE- SC707 40

Binary Mask: Round and Annular Illumination

•

Small

and k 1 >0.5

All power is inside pupil (for 0 th and



1 st orders)

•

Coherent source points have 3 beam interaction

• Larger s and k 1 <0.5

Some power is inside pupil (center of



1st orders is outside)

•

Coherent source points have 2 beam interaction Conventional or Circular Source Annular Source

Milster, Socha, Brooker SPIE- SC707 41

•

Binary Mask: 3-Beam Imaging

Let’s do an experiment

–

Set up the bench with:

•

L(25µm) Pitch grating

•

PH Source

–

Observe the behavior (position and contrast) of the image as the observation plane is moved from the perfect focus. Write down your observations.

–

What happens as the observation plane is moved beyond the point of zero contrast?

Milster, Socha, Brooker SPIE- SC707 42

Binary Mask: 3-Beam Imaging

–

Do you see reversed-contrast lines?

– – –

This type of focus behavior is indicative of

three-beam imaging

, where all of the power from the 0 and +/- 1 st diffraction orders passes the stop. Every point in the image is derived from three conjugate source points in the pupil.

Three-beam imaging has the characteristic that reversed-contrast planes can occur if the focus is too far or the resist is too thick.

Milster, Socha, Brooker SPIE- SC707 43

Binary Mask,



=248nm, NA=0.63, 300nm L/S

3-Beam Imaging

Milster, Socha, Brooker SPIE- SC707 44

•

Missing Orders

– – – – – –

Let’s do an experiment

–

Set the bench with

• •

L(25 µm) pitch PH source Draw a sketch of the image on the next page.

Block the zero diffraction order at the stop.

Draw a sketch of the image on the next page.

Does the pitch of the image change?

This type of focus behavior is indicative of

two-beam imaging

Every point in the image is derived from two conjugate source points in the pupil, which are widely separated and lead to a double-frequency image.

–

Now change the system to block either the +1 or -1 order, but let the zero order pass the stop.

– –

Draw a sketch of the image on the next page.

Observe the image pitch and defocus behavior. Write down your observations.

Milster, Socha, Brooker SPIE- SC707 45

Missing Orders

L pitch and PH source Block zero order Block ± 1 order Milster, Socha, Brooker SPIE- SC707 46

Binary Mask,



=248nm, NA=0.63, 250nm L/S

Milster, Socha, Brooker SPIE- SC707 47

  2(

 1)

Etched depth d SiO 2 E-Field Diffraction Orders

-5 -3 -1 q +1 +3 +5

Lens/Pupil

Phase Mask

E +1 -1 Position Grating Equation:

sin( q )  2  (



Pitch

)

-1 st +1 st

Milster, Socha, Brooker SPIE- SC707 48

Pure Phase – Chromeless

d SiO 2 Etched depth E-Field



  2(

 1)

E +1

Diffraction Orders

-5 -3 -1 q +1 +3 +5

-1 Position Grating Equation:

sin( q )  

Pitch

Lens/Pupil -1 st +1 st

Milster, Socha, Brooker SPIE- SC707 49

•

Phase Mask

The phase mask produces no zero order

Pupil (stop) +3 +1

Strong Image Modulation

-1 No zero order is emitted from the phase mask.

-3 NA=sin( q Max ) q Max q Max pupil +1 pupil -1 +1

Coherent limit

min   2 

NA For alternating phase shift grating LW

min   4 

NA p min

is the minimum Cr pitch that is at the limit of resolution. Photoresist CD k 1 =1/4   1 

No Image just constant Irradiance

-1 Milster, Socha, Brooker SPIE- SC707 50

•

Phase Mask

Let’s do an experiment

–

Set the bench with:

•

12.5µm Pitch Phase Mask

• •

14.25mm Diameter stop (No Magnet) 3mm Diameter Source (

~ 0.3)

–

Observe the light pattern at the stop. How many diffraction orders do you see?

– – – –

Draw a sketch of image and the light pattern at the stop on the next page.

Note the relative brightness of the zero order and the +/-1 st orders. If needed, remove the grating to identify where the zero order occurs. Observe the line pattern at the observation plane. (Block the zero order if present) How does the image pitch compare to using a simple grating mask?

Milster, Socha, Brooker SPIE- SC707 51

–

Phase Mask

Change the observation plane location. How sensitive is the observation – plane location to focus changes?

– – –

The phase mask has no zero order, and it produces a

double-frequency pitch

in the aerial image compared to a binary mask.

The minimum pitch in the image is half the minimum pitch of a simple grating mask.

The phase-mask image is relatively insensitive to focus changes, due to the missing zero order.

Milster, Socha, Brooker SPIE- SC707 Draw a sketch of image.

Draw a sketch of light pattern at the stop.

Milster, Socha, Brooker SPIE- SC707 

=248nm, NA=0.63, sigma = 0.3

•

Off-Axis Illumination

Illumination source shapes that do not have axial intensity as usually known as off-axis sources

–

Examples are annular, quadrupole, and dipole

•

Off-axis illumination helps to enable k1<0.5 with binary masks

–

Reduction of on axis source reduces “DC” terms and enhances contrast

•

A conventional on axis small source Some Off-axis sources Annular 45



Quadrupole 0



Quadrupole y- dipole x- dipole

Milster, Socha, Brooker SPIE- SC707 54

•

Coherent Off-Axis Illumination and a Binary Mask

Orders shift relative to pupil

+3 pupil +1 0 -1

Image Modulation

NA=sin( q Max ) q Max q Max -3

For 1:1 grating

pupil 0 -1 pupil 0

“Incoherent” limit

min   2 

NA LW

min   4 

NA P min

is the minimum pitch that is at the limit of resolution. Photoresist CD k 1 =1/4   1 

No Image just constant Irradiance

-1 Milster, Socha, Brooker SPIE- SC707 55

Binary Mask with Annular Illumination

Resolution limit with 0<

<1 for a circular source

s outer •

No grating - just blank mask

s center s inner

0th order

•

Grating period at cut-off frequency

 

-1st +1st 0th order

  limit  2

(1  s

outer

) •

Grating period resolution limit at given

min  (1 s 

outer

) 

For 1:1 grating,

min  s 

outer

) 

Milster, Socha, Brooker SPIE- SC707

-1st 0th order +1st

•

Off-Axis Illumination with a Binary Mask

Let’s do an experiment

–

Set up the bench with: system for minimum

•

S(8.2µm) pitch mask

•

PH Source centered on axis

–

Observe the pattern at the stop. Draw the light pattern at the stop on the next page.

–

Do you see an image? Sketch the camera output on the next page.

–

Move the source until at least two orders pass through the stop. Draw the pattern at the stop and the image on the next page.

Milster, Socha, Brooker SPIE- SC707 57

Off-Axis Illumination with a Binary Mask

Draw a sketch with the centered source Draw a sketch with decentered source Milster, Socha, Brooker SPIE- SC707 Camera Output Camera Output 58

Milster, Socha, Brooker SPIE- SC707 

=248nm, NA=0.63, sigma = 0.3

Different cases for off axis, k1=0.25

• • • • • •

Assume off axis illumination Assume dense L/S k1=0.25

– – 

Center of n=0 and n=1 diff. orders are at edge of lens



CD = LW = 0.25*Lambda/NA For 248nm illumination, NA=0.63

–

CD = 0.5*248nm/0.63 = 98nm



100nm L/S give k1=0.25

For 193nm illumination, NA=0.93

–

CD = 0.25*193nm/0.93 = 52nm



50nm L/S give k1=0.25

For 193nm illumination, NA = 1.2

–

CD = 0.25*193/1.2 = 40.2nm



40nm L/S gives k1 = 0.25

• •

Current off-axis results.

Actually might want whole orders inside with sigma=0.3

Milster, Socha, Brooker SPIE- SC707 60

•

Summary

What have we learned?

–

The basic optical components of a lithography system are the source, condenser and imaging lens.

–

The size and shape of the source influence properties of the aerial image.

–

The stop of the system determines the maximum angle of diffraction orders that can pass to the image.

– –

It takes at least two diffraction orders passing the stop to form a line-space image.

By increasing

, we can change from coherent-like illumination to partially-coherent illumination.

–

Partially coherent illumination can allow higher pitch in the image at the expense of reduced contrast.

–

2-Beam and 3-Beam geometries have different focus characteristics.

–

By using a phase-shift mask, the zero order is eliminated and the first diffraction orders move closer to the center.

–

Off-axis illumination can produce a half-pitch image, but the contrast is lower than with a phase-shift mask.

Milster, Socha, Brooker SPIE- SC707 61

References

Introductory Articles SPIE Proceedings for Microlithography Journal of Microlithography, Microfabrication, and Microsystems (JM 3 ) – SPIE Press Industry Magazines

Microlithography World www.pennwell.com

Books Intro to Fourier Optics and Statistical Optics by J. Goodman Resolution Enhancement Techniques and Optical Imaging in Projection Microlithography Alfred Wong, SPIE Press

Microlithography: Science and Technology

Ed: James Sheats and Bruce Smith Pub: Marcel Dekker

Linear Systems

by J. Gaskill

Milster, Socha, Brooker SPIE- SC707 62

References

Intro Papers “Using location of diffraction orders to predict performance of future scanners”, Peter Brooker Publication: Proc. SPIE Vol. 5256, p. 973-984, 23rd BACUS (2003) “Roles of NA, sigma, and lambda in low-k1 aerial image formation”, Peter D. Brooker Publication: Proc. SPIE Vol. 4346, p. 1575-1586, (2001) Advanced Papers U of A Dissertation by Doug Goodman (1979), Stationary Optical Projectors Papers by H.H. Hopkins for partial coherent imaging, Richards and Wolf for high NA

Milster, Socha, Brooker SPIE- SC707 63

Milster, Socha, Brooker SPIE- SC707

Thank You for Taking This Course!

Milster, Socha, Brooker SPIE- SC707

Backup Slides

•

Basic Illumination and Imaging

Pupil or “the aperture stop”: Physical Limiting aperture of system

–

Location and size defined by Chief Ray and Marginal Ray

• •

Chief Ray: Starts at edge of object (field) goes through center of pupil Marginal Ray: Starts at axial object and goes through edge of pupil Pupil Object

Chief ray

Marginal ray

q 

n’ = image side refractive index Aerial Image h’ n = object side refractive index

•

NA: numerical aperture

– –

defined by marginal ray maximum angle accepted by system

 

 sin q 

m h

 

NA NA



Milster, Socha, Brooker SPIE- SC707 66

•

Basic Illumination and Imaging

Imaging Lens

Let’s do an experiment

–

Calculate NA at the image plane for

r s =

_______

 

 sin q 

m NA

 sin q 

 sin tan 

r s

 1  

r s f

2     –

Calculate the coherent resolution limit in terms of pitch in the aerial image.

  min 



 

1

2   

________ 525 nm

’

Milster, Socha, Brooker SPIE- SC707 67

Optimum DOF and Modulation for Annular (and dipole)

-1 0 +1

NA * s

center

• 

/Pitch Optimum when phase differences between 0th and 1st orders are minimum

2 

 s

center

 

Pitch

center

 1 2 1 

Pitch NA

 1 1 4 

LW NA

Milster, Socha, Brooker SPIE- SC707 68

Optimum DOF and Modulation for Quadrupole

-1 0 +1

NA * s

center



/Pitch

•

Optimum when phase differences between 0th and 1st orders are minimum

2 

 s

center

2  

Pitch

center

 2 2 1 

Pitch NA

 4 2 1 

LW NA

Milster, Socha, Brooker SPIE- SC707 69

Off-axis Illumination Principles

Effects of different illumination modes

 Periodic features benefit most from QUASAR illumination  Optimum illumination is specific to reticle features 1.5

QUASAR hor. / vert.

conventional conventional 1.0

annular 0.5

annular QUASAR 45 ° lines Dense Lines @ 60% contrast QUASAR 0.0

0.0

0.5

1.0

Resolution [  /NA] 1.5

2.0

after IMEC 1997 Milster, Socha, Brooker SPIE- SC707 70

More Facts: Aerial Image Cross Section



=193nm, NA=0.75

Dense Lines vs.

(circular)

Image Intensity 0.50

Varying

s 1.3

Image Intensity 1.2

0.40

0.30

0.20

100nm L/S

0.1

0.3

0.5

0.7

0.9

0.5

0.4

0.3

0.2

0.1

0.0

1.1

1.0

0.9

0.8

0.7

0.6

150nm L/S

0.10

-100 -80 -60 -40 -20 0 20 40 Horizontal Position (nm) 60 80 100 -100 0 Horizontal Position (nm) 100 • • •

Increase sigma and contrast goes up (100nm L/S) Increase sigma and contrast goes down (200 nm L/S) Very confusing!!! What is going on??

Milster, Socha, Brooker SPIE- SC707 0.1

0.3

0.5

0.7

0.9

•

Lithography Imaging Laws

What limits the size of the photoresist pattern ?

–

Three paths to improve resolution:

• • •

Wavelength (



) Numerical Aperture (NA)

Photoresist CD

  1

k 1 : “Process” knob

–

Includes off-axis illumination, complex masks, high contrast photoresist, acid diffusion, etc…



• •

What limits the size of the optical (and/or aerial) image? (Assuming circular illumination source and binary reticle)

• • •



Partial Coherence or illumination (

)

– – s

=0: Coherent Limit

=1: Incoherent Limit Fine…but where do these come from??

Optical Resolution  1 2(1  s )  

•

Note: resolution is often written as Linewidth (LW) or critical dimension (CD) in the context with photoresist

Milster, Socha, Brooker SPIE- SC707 72

•

Basic Illumination and Imaging

Definition of Coherence Factor ‘Sigma’ Mask Plane Source Image Diameter Stop Diameter Source Pupil Edge (the “NA”) Condenser Imaging Lens

σ



Diameter of Source Image in Stop Diameter of Stop

Source Image View of Entrance Pupil with blank mask

If pupil diameter = NA, then source size = NA •

s (pupil or NA units) Milster, Socha, Brooker SPIE- SC707 73